10.2 The Occurrence of Critical Flow; Controls

Transcription

1 10. The Ourrene of Critial Flow; Controls In addition to the type of problem in whih both q and E are initially presribed; there is a problem whih is of pratial interest: Given a value of q, what fators determine the speifi energy E, and hene the depth y? Conversely, if E is given, what fators determine q? The answer to these questions is that there are many different kinds of ontrol mehanism whih an ditate "what depth must be for a given q, and vie versa". Example is the sluie gate; For a given opening of the gate there is a ertain relationship between q and the upstream depth, similarly for the downstream depth. Weirs and spillways are further examples of this kind of mehanism. The flow resistane due to the roughness of the hannel bed will have some effet. The flow situation in any hannel is substantially influened by the ontrol mehanisms operating within it. The notion of a "ontrol" - any feature whih determines a depth - disharge relationship - is of primary importane in the stu of free surfae flow. There are ertain features in hannel whih tend to produe ritial flow, and are therefore ontrols (see box) of a rather speial kind. Three types of ontrols namely (i) downstream ontrol (ii) upstream ontrol and (iii) Artifiial ontrol. are identified. Normally, the sub ritial flow deals with downstream ontrol and superritial flow deals with the upstream ontrol. The nature of these features, are determined by onsidering the general problem of flow without losses in a retangular hannel setion of onstant width, whose bed level may vary. This is a partiular situation of the transition problem. (See box).

2 Transition (flow basis): 1. Sub ritial to Sub ritial. Sub ritial to Super ritial (Hydrauli drop). Super ritial to Sub ritial (Hydrauli Jump) 4. Super ritial to Super ritial Transition Struture: Converging Diverging 1. Retangular ross setion to Retangular ross setion. Retangular ross setion to Trapezoidal ross setion. Trapezoidal ross setion to Trapezoidal ross setion 4. Trapezoidal ross setion to Retangular ross setion 5. Trapezoidal ross setion to irular ross setion or Horse shoe tunnel 6. Horse shoe tunnel to Trapezoidal ross setion 7. Horse shoe tunnel to Retangular ross setion et. Method of onnetion in transition (gradual): a. By straight wall b. By Quadrant (ylindrial). By warped The transition ould be abrupt suh as sudden expansion or sudden ontration. The transition ould be gradual over ertain distane. The transition an be in vertial plane suh as steps, humps, drops. The transition ould be both in plan and in elevation. Generally, transition are provided as inlet and outlet struture. The flow in suh transitions is three dimensional and omplex.

3 Following assumptions are made 1. Constant retangular hannel. Short reah.. No fritional loss. 4. Hydrostati pressure distribution is assumed. V g V g T.E y z y x z = f(x) Longitudinal setion b q Plan

4 The total energy H and q Q = = disharge per unit width b are onstant, ( ) q H = y+z+ = E+z = onstant gy differentiating with respet to x, the distae along the hannel de dz + =0 whih may be rewritten as de dz + = 0 Substituting and simplifying dz 1-F + = 0 de V ( =1-F ; F = ) gy V E = y + g - de d Q Q da = 1+ = 1+ A ga g de Q =1- T = 1- F ga (i.e) F QT = ga It is to be noted that the Froude number F plays a key role in this equation. This equation demonstrates in nutshell from a result from the E-y urve. If there is an upward step in the hannel bed, i.e., if dz/ is positive, then the produt ( 1-F ) must be negative and vie versa (see box).

5 If dz is positive Bed ( ) 1 F = negative F < 1 ( Subritial ) -ve ( depth dereases along x) F > 1 Superritial +ve depth inreases along x ( ) ( ) z If dz is negative Bed ( ) z 1 F = positive F < 1 ( Subritial ) +ve ( depth inreases along x) F > 1 Superritial -ve depth dereases along x ( ) ( ) However, if the hannel bed is horizontal i.e., dz 0 =,. Then the produt ( ) 1-F is then equal to zero. Hene, either = 0 or F = 1 (ritial flow). The first situation ours in the step-transition problem when dz 0 = upstream of the step and over the step, and in both ases F 1., 0 = both For the seond situation, the question is " Can a situation be visualized in whih dz 0 = and 0?

6 The answer is yes. Consider the Free outflow from a Lake as an example of ritial flow. Flow E o P y = E 0 dz = 0, 0, F = 1 An example of Critial - Free Outflow from a Lake When water is released from a lake over a short (but smooth) rest suh that it flows downstream freely. In other words either a free overfall within a short distane downstream or a steep slope whose bed resistane imposes no effetive onstraint on the flow. At the rest P, dz 0 = the flow is aelerating at this point, resulting in 0. Then the Froude number must be equal to unity, and hene the flow would be ritial. In ases of a sharp- edged (e.g., V noth weir) rest, and a ompletely free overfall, are onsidered as pressure distribution would be non hydrostati; for the reason the urvature will not be large. However, even if the vertial aelerations is large, as near brink of a free overfall, the flow is still an be approximated as the ritial ondition. Experimental evidene indiates that the flow depth right at the brink of an overfall is approximately 5 y, (i.e y ) and that y = y 7 at a distane upstream from the overall edge of weir of infinite height, the disharge is remarkably lose to that obtained by assuming ritial flow at the rest, despite the pronouned vertial urvature of the flow. Assuming that

7 the pressure distribution is hydrostati, it an be onluded that when water is released from a lake without any downstream onstraint ritial flow ours at the setion of maximum vertial onstrition: suh a setion is therefore a ontrol. Similarly that ritial flow ours at a orresponding horizontal onstrition. Free overfall over a sharp rested weir Free overfall (drop) Free Over fall over an arh dam End Depth or Brink Depth When the hannel terminates abruptly the end weir is known as The Weir of Zero height". The flow in the end reah of the hannel beomes an overfall. Measuring the depth at the end setion of the hannel, the disharge an be estimated. Rouse first identified this aspet in a horizontal retangular hannel (with sub ritial approah flow). The end depth (also alled the brink depth) was times the ritial depth.

8 When the anal drops suddenly, a free overfall is formed, sine flow hanges to superritial flow an be used as a measuring devie. v g y y H1 y y b - - Level Minimum drop distane Free overfall profile X y The drop distane should be more than 0.6y. Brink depth y b will be different at the entre and sides of the anal (whih is higher). The roughness of the anal affets the brink depth and hene the bed and sides should be finished smooth. H q = y + α gy o Differentiating w.r.t 'y' assuming Q to be onstant. dh o q = 1 α gy dh o 0 = if the flow is ritial, hene αq y = g If α =1, then Q = b g y Rouse showed y b / = y / yb Thus Q = b g This derivation is assumed for a free fall with an unonfined nappe. This value is modified as when the flow is two dimensional. This results in a error of to % respetively for the above two ases. The width of the anal should not be less than y. This is appliable to anals with slopes upto

9 P Thin weir plate - Free over fall y x Brink depth or End depth (y b) y yb L 1.4, x = to 4 y Brink depth

10 10.. Constrition in bed width In ase of a horizontal hannel bed and a variable width b, the energy equation an be written, taking z as a onstant but q as a variable funtion of x as V Total energy = TE = H = z + y + ( α = 1.0 ) g Q q.b = > H = z + y + = z + y + ga g.b y ( ) q x H = y + z + gy Differentiating both sides with respet to " x ", dh dz d ( ) q x = + + = 0 gy dh dz If = 0 and = 0 No energy loss, Horizontal hannel q q - + gy gy ( ) dq = 0 and by ontinuity equation q b = a onstant, Q. Then dq dq db = 0 = b + q = 0 dq db b = q dq Eliminating, between above two equations then it may be written as q q db ( 1-F ) - = 0 gy b ( ) y db i.e., 1-F - F = 0 b It an be onluded that ritial flow ours when db, i.e., at a setion of maximum horizontal onstrition. The ritial flow will not our at a setion of maximum width, but only at a setion of minimum width.

11 Converging db () i < 0 F<1 subritial then < 0 depth dereases as x inreases F>1 superritial then > 0 depth inreases as x inreases Diverging db i > 0 F<1 subritial then > 0 depth inreases as x inreases F>1 superritial then 0 depth inreases as x inreases < () Converging hannel F<1 < 0 Sub ritial Diverging hannel db < 0 F>1 > 0 Super ritial F<1 db F>1 Sub ritial > 0 Super ritial Horizontal onstrition

12 Derive the following equation for a non prismati hannel, assuming no energy loss. = S ο + y by y 1 y db. Solution: Total energy at any setion is given by V H = z + y + ( α = 1.0 ) g Differentiating wrt "x", dh dz d V = + + > (1) g dh But = Sf dz dh Similarly = S ο But S f = 0 => = 0 Substituting in eq: ( 1 ), d V 0 = - S ο + g + > () d Q 0 = - S ο + + ga Consider a retangular hannel with varying width d Q d Q Q d 1 = = ga g b y g b y Q db = g b y b y Q db Q = gb y gb y

13 Substituting this expresstion in eq: ( ), - S ο Q db Q + = 0 gb y gb y q q db => - Sο + 1 = 0 gy gb y But q g = y y db Sο +. by = y 1 : y